Question

A satellite moves around the earth in a circular orbit with speed $ v $ . If $ m $ is the mass of the satellite, its total energy is
(A) $ - \dfrac{1}{2}m{v^2} $
(B) $ \dfrac{1}{2}m{v^2} $
(C) $ - \dfrac{3}{2}m{v^2} $
(D) $ - \dfrac{1}{4}m{v^2} $

Answer 1

Hint: The total energy is the sum of the kinetic energy plus potential energy. Represent the potential energy in terms of velocity.

Formula used: In this solution we will be using the following formulae;
 $ PE = - \dfrac{{GMm}}{r} $ where $ PE $ is the potential energy, $ G $ is the universal gravitational constant, $ M $ is the mass of the earth, $ m $ is the mass of the satellite, and $ r $ is the radius of the earth.
 $ KE = \dfrac{1}{2}m{v^2} $ where $ KE $ is the kinetic energy, $ m $ is the mass of the satellite, and $ v $ is the speed of the satellite.

Complete Answer:
Generally, for an orbiting satellite about the earth, the total energy is given by the sum of the potential energy (which is due to its height above the planet) and the kinetic energy, which is due to its speed.
The potential energy can be given as
 $ PE = - \dfrac{{GMm}}{r} $ where $ G $ is the universal gravitational constant, $ M $ is the mass of the earth, $ m $ is the mass of the satellite, and $ r $ is the radius of the earth.
 Whereas the kinetic energy can be given as
 $ KE = \dfrac{1}{2}m{v^2} $ where $ m $ is the mass of the satellite, and $ v $ is the speed of the satellite.
We should add them together. But before that we shall express the potential energy in terms of velocity.
To do so, we Note that the centripetal force required for the orbit is provided by the force of attraction, i.e.
 $ \dfrac{{GMm}}{{{r^2}}} = \dfrac{{m{v^2}}}{r} = PE $
Hence, total energy is
 $ TE = PE + KE = = - m{v^2} + \dfrac{1}{2}m{v^2} $
 $ \Rightarrow TE = - \dfrac{1}{2}m{v^2} $
Hence, the correct option is A.

Note:
Alternatively, recall that the total energy of a satellite in orbit can be given as
 $ TE = - \dfrac{{GMm}}{{2r}} $
From above, we see that
 $ - \dfrac{{GMm}}{r} = - m{v^2} $
Hence by dividing both sides by 2,
 $ PE = - \dfrac{{GMm}}{{2r}} = \dfrac{{m{v^2}}}{2} = - \dfrac{1}{2}m{v^2} $ .